If the distance between the foci of an ellips | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given that the distance between the foci of an ellipse is equal to the length of the latus rectum.The distance between the foci is $2ae$ and the l

Vedclass · Accepted Answer

$\frac{1}{2}(\sqrt{5}-1)$

Vedclass · Answer

(C) Given that the distance between the foci of an ellipse is equal to the length of the latus rectum.The distance between the foci is $2ae$ and the length of the latus rectum is $\frac{2b^2}{a}$.Therefore, $2ae = \frac{2b^2}{a} \implies ae = \frac{b^2}{a} \implies b^2 = a^2e$.We know that for an ellipse, $b^2 = a^2(1 - e^2)$.Substituting $b^2 = a^2e$ into the equation, we get $a^2e = a^2(1 - e^2)$.Dividing by $a^2$ (since $a 
eq 0$), we get $e = 1 - e^2$.Rearranging gives the quadratic equation $e^2 + e - 1 = 0$.Using the quadratic formula $e = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, we get $e = \frac{-1 \pm \sqrt{1^2 - 4(1)(-1)}}{2(1)} = \frac{-1 \pm \sqrt{5}}{2}$.Since eccentricity $e$ must be positive $(0 < e < 1)$, we take the positive root: $e = \frac{\sqrt{5} - 1}{2}$.

If the distance between the foci of an ellipse is equal to the length of the latus rectum, then its eccentricity is

Similar Questions

If the distance between the foci of an ellipse is $6$ and the length of the minor axis is $8$,then the eccentricity is

If the lines $2x - y + 3 = 0$ and $4x + ky + 3 = 0$ are conjugate with respect to the ellipse $5x^2 + 6y^2 - 15 = 0$,then $k$ equals:

If an ellipse with foci at $(3,3)$ and $(-4,4)$ is passing through the origin,then the eccentricity of that ellipse is

The equation of an ellipse,whose vertices are $(2, -2)$ and $(2, 4)$ and eccentricity is $\frac{1}{3}$,is

The distance between the directrices of the ellipse $\frac{x^2}{36}+\frac{y^2}{20}=1$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module